AC motor: what capacitance corrects power factor to 1.0?

[RagincajunLA]

Hey guys I just need some help on something

The question is: A motor attached to a 120 V/60 Hz power line draws an 8.50 A current. Its average energy dissipation is 800 W.

I got the first 3 parts of this question correct...

1. What is the power factor
ans = .784

2. What is the rms resistor voltage?
ans = 94.1 V

3. What is the motor's resistance?
ans = 11.1 ohms

now all of these answers are right but i can't seem to get the last one and it is...
4. How much series capacitance needs to be added to increase the power factor to 1.0?

 

[K^2]

I'm assuming 120V and 8.5A are also RMS. It's been a while, but it's the only thing that makes sense with 800W. If either one is peak-to-peak, that'd give you power factor that's greater than 1.0 and that's nonsense. Anyways...

You know the resistance. You know the total magnitude of the impedance, because you have voltage to current ratio. So you can compute inductive component of the impedance. Now all you have to do is find the capacitance such that capacitive impedance at 60Hz cancels your inductive impedance. Then the net impedance of the motor is just the resistance, and your power factor becomes 1.0.